Question

Let $S=\left\{\left(x,y\right)|x^2+y^2\le 1,x\ge 0\right\}$

Find the centroid of S.

Short answer

Answer is$\left(\frac{4}{3}\mathrm{\pi },0\right)$

Step-by-step solution

Step 1. Given information

$S=\left\{\left(x,y\right)|x^2+y^2\le 1,x\ge 0\right\}$

Step 2. Explanation

$S=\left\{\left(x,y\right)|x^2+y^2\le 1,x\ge 0\right\}$

$\begin{array}{rcl}\overline{x}& =& \frac{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}xdydx}{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dydx}\\ \overline{x}& =& \frac{{\int }_0^{1}x[y{]}_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx}{{\int }_0^1[y{]}_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx}\\ & =& \frac{2{\int }_0^{1}x\sqrt{1-x^2}dx}{2{\int }_0^1\sqrt{1-x^2}dx}\\ & =& \frac{{\int }_0^{1}x\sqrt{1-x^2}}{{\int }_0^1\sqrt{1-x^2}dx}\\ & =& \frac{{\left[-\frac{1}{3}{\left(1-x^2\right)}^{3/2}\right]}_0^1}{{\left[\frac{1}{2}\left\{x\sqrt{1-x^2}+{\sin }^{-1}x\right\}\right]}_0^1}\\ & =& \frac{\left(\frac{1}{3}\right)}{\left(\frac{\pi }{4}\right)}\\ & =& \frac{4}{3}\mathrm{\pi }\end{array}$

$\begin{array}{rcl}\overline{y}& =& \frac{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ydydx}{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dydx}\\ & =& \frac{{\int }_0^1{\left[\frac{y^2}{2}\right]}_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx}{{\int }_0^1[y{]}_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx}\\ & =& \frac{{\int }_0^1\left[0\right]dx}{{\int }_0^1\left[0\right]dx}\\ & =& 0\end{array}$